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    ITS really annoying, i can do partial fractions just fine and integrate fraction just fine but these questionS!! I keep getting part of the answer or like in a different form!!!

    I'll give you a sample question and my answer to it and the textbook answer to it.

    Question: find integral of : 2x/3-x (using partial fractions)

    My answer : -6ln(3-x) +c

    Textbook answer : -2x -6ln(3-x) +c

    help quick pleassssssssssssseee.


    Thanks everyone!
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    Well, what partial fraction representation did you get?
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    I've got the same answer as the textbook.
    Looks like you've done the partial fraction incorrectly.
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    (Original post by watchthis)
    I've got the same answer as the textbook.
    Looks like you've done the partial fraction incorrectly.
    Show me please! Your working! So i can stop hyperventilating!
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    (Original post by DFranklin)
    Well, what partial fraction representation did you get?
    -3/3 + 2/1- (a third x)

    oh wait!!!!!!!! i see it!
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    (Original post by Destroyviruses)

    Question: find integral of : 2x/3-x (using partial fractions)


    help quick pleassssssssssssseee.
    Surely it has a linear denominator already...where do partial fractions come into it?
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    (Original post by Destroyviruses)
    ITS really annoying, i can do partial fractions just fine and integrate fraction just fine but these questionS!! I keep getting part of the answer or like in a different form!!!

    I'll give you a sample question and my answer to it and the textbook answer to it.

    Question: find integral of : 2x/3-x (using partial fractions)

    My answer : -6ln(3-x) +c

    Textbook answer : -2x -6ln(3-x) +c

    help quick pleassssssssssssseee.
    Well get the partial fraction first, then replace the intergral with the partial fractions version and intergrate that
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    lol.
    well, the fraction 2x/3-x, because it is a mixed fraction or really a top heavy fraction you'll have to convert it into the from A + B/3-x
    then you put A + B/3-x = 2x/3-x
    Then multiply everything by 3-x, you get A(3-x) + B = 2x
    Then you can find A and B and you're sorted.
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    (Original post by watchthis)
    I've got the same answer as the textbook.
    Looks like you've done the partial fraction incorrectly.
    Nevermind got it!!!!!!!!
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    I can confirm, I got the answer in the book. Remember when doing partial fractions to divide the original fraction if the "degree" in the numerator is the same as the degree in the denominator. I.e. in this case it's x on top and bottom. To do partial fractions the degree of the denominator must be bigger than the degree of the numerator like having x in the numerator and x^2 in the denominator.

    I hope that makes some sense. If you do that and get the correct partial fractions, it's a piece of cake.
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    (Original post by watchthis)
    lol.
    well, the fraction 2x/3-x, because it is a mixed fraction or really a top heavy fraction you'll have to convert it into the from A + B/3-x
    then you put A + B/3-x = 2x/3-x
    Then multiply everything by 3-x, you get A(3-x) + B = 2x
    Then you can find A and B and you're sorted.
    I love you!
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    (Original post by watchthis)
    lol.
    well, the fraction 2x/3-x, because it is a mixed fraction or really a top heavy fraction you'll have to convert it into the from A + B/3-x
    then you put A + B/3-x = 2x/3-x
    Then multiply everything by 3-x, you get A(3-x) + B = 2x
    Then you can find A and B and you're sorted.
    Lol guess what i did A/3 + B/ 1- athirdx
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    So what's it got to do with partial fractions? Surely it's just a very basic algebraic division, and a basic integration.
 
 
 
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